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Q11:
Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term?

Solution :

Given: An Arithmetic Progression (AP) defined by the sequence: $3, 15, 27, 39, \dots$

To Find: The term number $n$ such that the $n^{th}$ term ($a_n$) is $132$ more than the $54^{th}$ term ($a_{54}$).

Step 1: Identify the parameters of the Arithmetic Progression.

The general form of an AP is $a, a+d, a+2d, \dots$, where $a$ is the first term and $d$ is the common difference.

From the given sequence $3, 15, 27, 39, \dots$:

First term ($a$) = $3$

Common difference ($d$) = $a_2 - a_1 = 15 - 3 = 12$

[Verification: $27 - 15 = 12$ and $39 - 27 = 12$. Since the difference is constant, $d = 12$.]

Step 2: State the formula for the $n^{th}$ term of an AP.

The formula for the $n^{th}$ term of an AP is given by:

$a_n = a + (n - 1)d$

Step 3: Calculate the $54^{th}$ term ($a_{54}$).

Using the formula $a_n = a + (n - 1)d$ with $n = 54$, $a = 3$, and $d = 12$:

$a_{54} = 3 + (54 - 1) \times 12$

$a_{54} = 3 + (53 \times 12)$

$a_{54} = 3 + 636$

$a_{54} = 639$

Step 4: Formulate the equation based on the problem statement.

The problem states that the $n^{th}$ term is $132$ more than the $54^{th}$ term:

$a_n = a_{54} + 132$

Substituting the value of $a_{54}$ found in Step 3:

$a_n = 639 + 132$

$a_n = 771$

Step 5: Solve for $n$ using the $n^{th}$ term formula.

Substitute $a_n = 771$, $a = 3$, and $d = 12$ into the general formula $a_n = a + (n - 1)d$:

$771 = 3 + (n - 1) \times 12$

Subtract $3$ from both sides:

$771 - 3 = (n - 1) \times 12$

$768 = (n - 1) \times 12$

Divide both sides by $12$:

$\frac{768}{12} = n - 1$

$64 = n - 1$

Add $1$ to both sides:

$n = 64 + 1$

$n = 65$

Final Answer: The 65th term of the AP will be 132 more than its 54th term.


More Questions from Class 10 Mathematics Arithmetic Progression EXERCISE 5.2


CBSE Solutions for Class 10 Mathematics Arithmetic Progression


Chapters in CBSE - Class 10 Mathematics


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